Let O be the center of the circle, AB be the chord and PQ be the perpendicular bisector of AB which intersects AB at point M.
The perpendicular bisector of the chord AB always passes through the center O.
Join PA and PB so that we have two triangles PAM and PBM.
In ΔPAM and ΔPBM:
AM = BM (M is the midpoint of AB ∵ perpendicular
bisector of AB intersects at M)
∠PMA = ∠PMB = 90°
PM = PM (common)
∴ ΔPAM ≅ ΔPBM (SAS congruence rule)
∴ AP = BP
⇒ arc PXA = arc PYB (by CPCT)
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